Energy Loss of Particles in Matte

The Energy Loss of Particles in Matter
I. Attenuation Length of a Photon Beam in a thin Absorber
A beam of photons will be attenuated in the forward direction in a material of thickness x based on simple
scattering or absorption processes. The linear attenuation coefficient µ scales with material density ρ, and
the mass attenuation coefficient µ/ ρ is often reported.
! dI
I =!("N A
)# !dx =!µ!dx!!
A
!!!!!!!!!!!!!!I = I 0
!e ! µX !!!!!!
albedo or
back scatter
I 0
Secondary
α,β,γ
processes
I
Part of the beam may be absorbed in the material or scattered out of the beam. Sometimes the beam may
create secondary scattering processes and new forms of α, β ,γ radiation may emerge from the absorber.
II. Energy loss by Heavy Charged Particle (Stopping Power)
Heavy charged particles ( charge z) loose energy principally be ionization processes in materials. The
ionization cross section is denoted by σ ionization and measured in barnes (10 -24 cm 2 ) The Bethe Bloch
atoms
!/cm 3
formula gives the energy dependence of the energy loss. The electron density n e
= !N A
A
Alpha particles ( z=2) will loose four times as much energy in matter as protons (z=1).
! dE
dx = 4" z 2 % e 2
mc 2 # 2 &
' 4"$ 0
x
2
( %
)
*
n e
+ln 2mc2 # 2 (
&
'
I(1 ! # 2 ))
* !# 2 Bethe ! Bloch Formula
" Z
http://pdg.lbl.gov/2009/AtomicNuclearProperties/index.html

III. Range
We can determine the mean range of of heavy charged particles of kinetic energy T by integration of the
stopping power -dE/dx.
R(T ) =
T
'
0
! dE $
"
#
dx %
&
(1
dE
IV. Range of Alpha Particles in Air
The range of alpha particles in air can be parameterized as
R = 0.4 E Eα < 4MeV
R= 1.24E – 2.62 4 < Eα

e
e’
e
e’
e ion
hf
Ionization
Radiation
The radiation yield is the fraction of the total dE/dx producing radiative photons.
(! dE dx) Y =!!
rad
(! dE dx) + ! dE
rad dx
( ) ion
! 6 " 10!4 ZT
!!#!!Radiation!Yield of stopped electron
1+ 6 " 10 !4 ZT
The radiation yield of an absorber can be a dangerous source of secondary radiation! The number of
gamma rays produced falls as 1/Eγ but higher energy photons at the end of the spectrum will escape the
absorber. The radiation flows in the direction of the incident electron beam with a Δθ∼1/ E γ distribution .
dN !
dE
Infrared
photons
E = Ε’ + Eγ
Eγ
E γ max= Ε ε
E
E’
E γ
IVa. X-rays production by electrons
X-rays are copiously produced by stopping electrons in a heavy metal target of high-Z material. The
generated x-ray spectra is composed of a (a) continuous bremsstrahlung portion and (b) an x-ray line
spectra described by the Bohr model. Zeff takes in to account the effective shielding of the nucleus
by inner oribital electrons and other spin-orbit and spin-spin effects.
" d! %
#
$
dk &
'
Brem
"
~!
#
$
Z 2
T i
%
&
' ! 1 k !ln " 4T i %
#
$
k &
' !!!!!!!!!!!!!!!!(a)
E L(
!!Z eff 2 ) 13.6eV !(! 1
n 1
2 * 1
n 2
2 )!!!!!!!!!!!!(b)
electron
beam
x-rays
Metal
target
In a typical electron x-ray tube, electrons are accelerated to the target by Vacc = 10’s of KVolts of
potential. A 100KV x-ray tube will produce at best 10 KeV x-rays !

The bremsstrahlung is caused by the de-acceleration of the electron in the metal. The x-ray lines are from
the filling of inner core electrons that have been displaced.
K,L,M x-rays are produced in materials when inner atomic electrons are dislodged or ionized. Following
the Bohr model, Mosely predicted x-rays would follow the pattern:
E K!
!!(Z " 1) 2 # 13.6eV !(! 1 1 2 " 1 2 2 )!
E K$
!!(Z " 6) 2 # 13.6eV !(! 1 1 2 " 1 3 2 )!
n=5
n=4
E L!
!!(Z " 6) 2 # 13.6eV !(! 1 2 2 " 1 3 2 )!
E L$
!!(Z " 6) 2 # 13.6eV !(! 1 2 2 " 1 4 2 )!
etc.
K α K β K γ
Computed Tomography
The attenuation of X-rays passing through an object can be used for imaging – computed tomography. For
a monoenergetic x-ray beam and a variable density absorber we can integrate over the line of sight of the
beam and obtain the image after many measurements are made. A typical scenario of a CT scan is shown
below.
I x,y
= I 0
!e ! " µ(x,y)! s source
!!!!!
ln(I 0
/ I x,y
)!=!"
µ(x, y)ds
!1
µ(x, y) =!! #
$ ln(I 0
/ I x,y
)!%
& '!x-ray!image
n=3
n=2
n=1
L α L β L γ
M α M β M γ
detector

Questions to Answer:
1) Estimate the kinetic energy at which collisional and radiative stopping power are equal for electrons in
(a) Be , (b) Cu, (c) Pb.
2) Estimate the fraction of the energy of a 2 MeV beta ray that is converted in to bremsstrahlung when the
particle is absorbed in lead.
3) Use Table 6.1 to estimate the range in cm in air for electrons of (a) 50keV, (b) 830 keV, (c) 100 MeV
4) Use Table 5.3 to determine the minimum energy that a proton must have to penetrate 30cm of tissue, the
approximate thickness of the human body.
5) Using table 5.3 for protons mass stopping power of water, estimate the stopping power of Lucite
(ρ=1.19g/cc) for a 35 MeV proton.
6) How far will 10 MeV alphas penetrate in air (ρ=1.24 x 10 -3 g/cc, STP)
7) Estimate the energy of the Kα X-ray in Pb.